C
CraftExam
high importancemedium8 min

Integration by Partial Fractions

Integration by partial fractions decomposes a proper rational function into simpler fractions whose integrals are standard, usually logarithmic or inverse trigonometric.

Practice This Concept

Learn the concept

Student-friendly explanation

This method applies to rational functions where the numerator degree is less than the denominator degree after any necessary division. The denominator is factorised, unknown constants are found, and each simpler fraction is integrated separately.

How to write this in exams

  1. 1

    Start with the exact idea

    Integration by partial fractions decomposes a proper rational function into simpler fractions whose integrals are standard, usually logarithmic or inverse trigonometric.

  2. 2

    Then show how to use it

    Check degrees of numerator and denominator. If needed, divide first. Factorise the denominator. Write the correct partial fraction form. Find constants by substitution or coefficient comparison. Integrate each term. Combine logarithms only if it simplifies cleanly.

  3. 3

    Add one concrete example

    1/[(x-1)(x+2)] can be written as A/(x-1)+B/(x+2), then A and B are found before integrating.

  4. 4

    Avoid this incomplete answer

    Writing A/(x+1)+B/(x+2) correctly but solving constants with sign errors, leading to ln|x+1|+ln|x+2| instead of a difference.

Definition

Integration by partial fractions decomposes a proper rational function into simpler fractions whose integrals are standard, usually logarithmic or inverse trigonometric.

Example

1/[(x-1)(x+2)] can be written as A/(x-1)+B/(x+2), then A and B are found before integrating.

Rule to remember

Method condition: use partial fractions for rational functions P(x)/Q(x), after ensuring degree P < degree Q. For distinct linear factors, write A/(x-a)+B/(x-b). For repeated factors, include A/(x-a)+B/(x-a)^2 and so on. Then integrate ∫dx/(x-a)=ln|x-a|+C.

Memory hook

Break the rational expression first; integrate only after the pieces are simple.

Examples and method

Worked example

Evaluate ∫dx/[(x+1)(x+2)]. Write 1/[(x+1)(x+2)]=A/(x+1)+B/(x+2). Then 1=A(x+2)+B(x+1). Put x=-1: 1=A, put x=-2: 1=-B, so B=-1. Integral =∫[1/(x+1)-1/(x+2)]dx=ln|x+1|-ln|x+2|+C=ln|(x+1)/(x+2)|+C.

Method to apply

Check degrees of numerator and denominator. If needed, divide first. Factorise the denominator. Write the correct partial fraction form. Find constants by substitution or coefficient comparison. Integrate each term. Combine logarithms only if it simplifies cleanly.

Diagram support

No diagram is required. A factor table or coefficient comparison layout is useful for organizing constants.

How CBSE asks it

Usually appears as 3-mark or 5-mark questions with distinct linear factors, repeated linear factors, or a rational function needing division first.

Avoid common mistakes

Common confusion

Students often skip checking whether the fraction is proper. If the numerator degree is greater than or equal to the denominator degree, polynomial division must come first.

Common wrong answer

Writing A/(x+1)+B/(x+2) correctly but solving constants with sign errors, leading to ln|x+1|+ln|x+2| instead of a difference.

Exam tip

After finding constants in the decomposition, recombine mentally or by substitution of easy x-values to check before integrating.

Quick check

Decompose 1/[(x+1)(x+2)] into partial fractions.

1/[(x+1)(x+2)]=1/(x+1)-1/(x+2).

Answer writing and exam use

1-mark answer

Integration by partial fractions decomposes a proper rational function into simpler fractions whose integrals are standard, usually logarithmic or inverse trigonometric.

2-mark answer

Integration by partial fractions decomposes a proper rational function into simpler fractions whose integrals are standard, usually logarithmic or inverse trigonometric. Method condition: use partial fractions for rational functions P(x)/Q(x), after ensuring degree P < degree Q. For distinct linear factors, write A/(x-a)+B/(x-b). For repeated factors, include A/(x-a)+B/(x-a)^2 and so on. Then integrate ∫dx/(x-a)=ln|x-a|+C. 1/[(x-1)(x+2)] can be written as A/(x-1)+B/(x+2), then A and B are found before integrating.

3-mark answer

This method applies to rational functions where the numerator degree is less than the denominator degree after any necessary division. The denominator is factorised, unknown constants are found, and each simpler fraction is integrated separately. Method condition: use partial fractions for rational functions P(x)/Q(x), after ensuring degree P < degree Q. For distinct linear factors, write A/(x-a)+B/(x-b). For repeated factors, include A/(x-a)+B/(x-a)^2 and so on. Then integrate ∫dx/(x-a)=ln|x-a|+C. Evaluate ∫dx/[(x+1)(x+2)]. Write 1/[(x+1)(x+2)]=A/(x+1)+B/(x+2). Then 1=A(x+2)+B(x+1). Put x=-1: 1=A, put x=-2: 1=-B, so B=-1. Integral =∫[1/(x+1)-1/(x+2)]dx=ln|x+1|-ln|x+2|+C=ln|(x+1)/(x+2)|+C. Usually appears as 3-mark or 5-mark questions with distinct linear factors, repeated linear factors, or a rational function needing division first. Writing A/(x+1)+B/(x+2) correctly but solving constants with sign errors, leading to ln|x+1|+ln|x+2| instead of a difference.
MCQ Quiz

Practice this concept with focused MCQs

Open the concept quiz intro first, review the test details, and then start a focused MCQ set from this concept only. Instant score and answer review are live now.

10 MCQs5 MinutesInstant Results
Practice This Concept

Help improve this page

Found something confusing, incorrect, or missing?

Practice This Concept